\( \DeclareMathOperator{cosec}{cosec} \)

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Here are some exam-style questions on this statement:

- "
*A continuous random variable \(X\) has probability density function \(f\) defined by*" ... more - "
*The continuous random vanable \(X\) has probability density function:*" ... more

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The **variance** of a discrete random variable measures the spread or dispersion of the probability distribution. Mathematically, if \( X \) is a discrete random variable with mean \( \mu \) and probability mass function \( p(x) \), the variance \( \sigma^2 \) is given by:
$$
\sigma^2 = \sum (x - \mu)^2 \cdot p(x)
$$

For **continuous random variables**, we use probability density functions (pdfs) instead of probability mass functions. The pdf, denoted by \( f(x) \), gives the likelihood of the variable taking on a particular value. The total area under the curve of a pdf is always 1. The **mode** of a continuous random variable is the value of \( x \) for which the pdf has its maximum value. The **median** is the value \( m \) such that the area under the curve to the left of \( m \) is 0.5.

The **mean** (or expected value) of both discrete and continuous random variables is a measure of the central tendency. For a discrete random variable \( X \) with probability mass function \( p(x) \), the mean \( \mu \) is:
$$
\mu = \sum x \cdot p(x)
$$
For a continuous random variable with pdf \( f(x) \), the mean is:
$$
\mu = \int_{-\infty}^{\infty} x \cdot f(x) \, dx
$$
The **variance** and **standard deviation** for continuous random variables can be found using similar formulas, but with integration instead of summation.

When we apply a **linear transformation** of the form \( Y = aX + b \) to a random variable \( X \), the mean and variance change in predictable ways. The new mean \( \mu_Y \) and variance \( \sigma_Y^2 \) are given by:
$$
\mu_Y = a\mu_X + b \\
\sigma_Y^2 = a^2\sigma_X^2
$$
The standard deviation \( \sigma_Y \) of \( Y \) is \( |a| \) times the standard deviation \( \sigma_X \) of \( X \).

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